PHYS463 Electricity& Magnetism III ( ) Solution #1

Transcription

1 PHYS463 Electricity& Magnetism III ( ) lution #. Problem 3., p.5: Find the average potential over a spherical surface of radius R due to a point charge located inside (same as discussed in 3..4, only with r<r). (in this case, Laplace s euation does not hold within the sphere.) Show that, in general, ave center + Q enc 4πε 0 R, where center is the potential at the center due to all the external charges, and Q enc is the total enclosed charge. lution: Assume a point charge inside the sphere (R >r). The distance from the charge to a point on the spherical surface is (law of cosines) R 2 + r 2 2rR cos θ The potential at that point on the surface is 4πε 0 4πε 0 R2 + r 2 2rR cos θ The average potential on the surface is ave I 4πR 2 4πε 0 4πR 2 4πε 0 ds (R 2 + r 2 2rR cos θ) /2 R 2 sin θdθdφ / lution

2 4πε 0 2rR 4πε 0 4πε 0 R R2 + r 2 2rR cos θπ ( r + R R r ) 2rR 4πε 0 R 0 [r + R (R r)] 4πε 0 2rR If there are more than one point charge within the sphere, the average potential is the superposition of the potentials produced by all charges ave X 4πε 0 R Q enc 4πε 0 R where Q enc is the total enclosed charge. Combine the result in 3..4 for charges outside the sphere: 0 ave center the total averaged potential is Q.E.. ave center + Q enc 4πε 0 R 2. Problem 3.3 The Laplace in spherical coordinates is 2 r 2! + sin ϑ! 2 + r 2 r r r 2 sin ϑ ϑ ϑ r 2 sin 2 ϑ φ 2 Since (r), the Laplace s euation becomes an ordinary.e. d r 2 d! r 2 dr dr This leads to or integrate once more r 2 d dr d r 2 d! dr dr c d dr c r 2 c r + b 2 / lution

3 In the cylindrical coordinate system 2 r! r r r + r 2 2 r z 2 Since (r), the Laplace s euation becomes an ordinary.e. r d dr r d! dr This leads to or integrate once more This leads to or integrate once more r d dr r 2 d dr d r d! dr dr c d dr c r r + b d r 2 d! dr dr c d dr c r 2 c r + b 3. Problem 3.4 Similar to the class notes (note), let s prove it by contradiction Assume φ and φ 2 are solutions to Poisson s euation and satisfy the boundary condition φ S,namely, 2 φ ρ and φ S φ S or φ φ ² 0 2 φ 2 ρ ² 0 and φ 2S φ S or φ 2 S S S φ S Consider the difference It obviously obeys φ d φ 2 φ 2 φ d 3 / lution

4 and φ ds or φ d Let s make up a vector function S φ 2 S φ S A φd φ d and apply Gauss s theorem to the integration over the closed surface,s I A ³ d S A d ( (φ d φ d )) d I LHS But the normal component A ns φ {z} ds φ ds or φ d 0 on the boundary. S RHS I A ns ds φ d φ ds {z} φ d S ds 0 on the boundary due to the fact that S ( (φ d φ d )) d φ d φ d + φ d 2 φ {z d d } φ d 2 d φ d 2 d Now, φ d 2 0and R φ d 2 d φ d 2 or φ d 0. Therefore, φ d C const. (everywhere in the volume ) φ d φ 2 φ + C Two solutions are eual except a physically insignificant constant C. The electric field is uniuely determined since E φ φ 2 E 2 4 / lution

5 Figure : 4. Problem 3.7 Note: should replace a and a should replace R in the textbook (a) The foollowing is known 0 a, x a2 (r, θ)! + 0! 4πε 0 r + r r + ar r r 2 2r cos θ r 2 + x 2 2x cos θ r r 2 + ³ 2 a 2 2 a 2 r cos θ a a 2 +(r/a) 2 2r cos θ The potential at that point is 2 + r 2 2r cos θ a 2 +(r/a) 2 2r cos θ At r a 2 + a 2 2a cos θ a 2 +(a/a) 2 2a cos θ 5 / lution

6 (b) E e r + e θ +! e z e r +! e θ r r θ z r r θ outer surface E r ra+ r cos θ ( + 2 +r 2 2r cos θ) 3 a cos θ ( + 2 +a 2 2a cos θ) 3 inner surfacee r ra 2 a 2 a( 2 +a 2 2a cos θ) 3 σ (θ) ² 0 ³ Er ra+ E r ra 4π r(/a) 2 cos θ ³ a 2 +(r/a) 2 2r cos θ a(/a) 2 cos θ ³ a 2 +(a/a) 2 2a cos θ 2 a a ³ 2 + a 2 2a cos θ 3 (c) Theforceonthepositivecharge due to the induced charge 0 ³ 0 F ( x) 2 ³ a ³ a 2 is 2 a2 ( 2 a 2 ) 2 based on the force between two point charges. The - sign indicating an attraction force. The work done by the external force (or the negative of the work done by the field) while the charge is moved from the infinite to is the energy of the system W F d l In vector form F a 2 ( 2 a 2 ) 2 e r, d l d e r W a2 a 2 F d l " # 2 ( 2 a 2 ) ( 2 a 2 ) 2 d a2 8π² 0 ( 2 a 2 ) ra 6 / lution

7 5. Problem 3.9. An infinitively long and thin wire with a uniform line charge density λ is placed above a grounded, infinitively large conducting plate at a height h as shown (The wire runs parallel to the y-axis). (a) Calculate the electric potential in the half space z>0. Image line current at (0,0,-h) will ensure φ onthez plane. The electric field produced by a line charge can be obtained using Gauss s law I E d A and the potential ² 0 E r (2πrl) λl E r ² 0 λ 2π² 0 r r E r dr λ r dr r 0 2π² 0 r 0 r (r) (r 0 ) λ µ r 2π² 0 r 0 Or select (r 0 )whenr 0 h in our case (r) λ µ r 2π² 0 h 7 / lution

8 Superposition of potential produced by line charge and image charge results φ(x, z) λ 2π² 0 r+ r! λ x 2 +(z h) 2! x 2 +(z + h) 2 for z>0 (b) erive the induced charge density on the conducting plate on the x y plane. φ σ (x) ² 0 E z ² 0 z z λ 2(z h) 2(z + h) 4π x 2 2! +(z h) x 2 +(z + h) 2 z λ h π x 2 + h 2 (c) Find the induced charge density per unit length (along the y-direction). Hint: dt +t arctan(t) 2 The charge on the strip y extending <x<+ is + σ (x) ydx The charge density per unit length is λ ind. + y σ (x) dx λ + h π x 2 + h dx 2 λ + π (x/h) 2 + d (x/h) λ + dt π +t λ 2 π arctan(t) + λ µ π π 2 π λ 2 (d) What is the potential in the half space z<0? 6. Problem 3-. φ(x, y, z) 0 for z<0 Let s locate image charge lines +λ and λ with infinitesimal thickness inside the negatively and positively charged transmission lines, respectively, so that the cylinder potential are constant Φ + and Φ on the two transmission lines. 8 / lution

9 Figure 2: d: istance between the centers of two transmission lines. : istance between the images. x: istance between image and the centre of the transmission line. In order to find the expression of x in terms of a and d, let s euate Φ A and Φ B. at point A: r +A a x, r A 2d x a at point B: r +B a + x, r B 2d x + a Recall the potential produced by two infinitesimally thin parallel transmission lines we discussed in the lecture: Φ(r +,r ) λ! r 2πε 0 r + () The euipotential condition (series of cylinders) is or r A r +A const. r B r +B 2d x a a x 2d x + a a + x 9 / lution

10 lve for x: x d ± d 2 a 2 (2) Since d>a,d+ d 2 a 2 >a,x d + d 2 a 2 represents a case with image charge outside the pipes in the region where the potential is considered. This case should be discarded. We choose x d d 2 a 2 (3) The potential at A is then: Φ A λ 2π² 0! 2d x a a x 2d x a d x + d a d 2 a 2 +(d a) a x d 2 a 2 (d a) 2d x a a x d2 a 2 +(d a) d2 a 2 (d a) h d2 a 2 +(d a) i 2 h d2 a 2 (d a) ih d2 a 2 +(d a) i d2 a 2 +2(d a) d 2 a 2 +(d a) 2 (d 2 a 2 ) (d a) 2 2(d a) d 2 a 2 +2d 2 2ad 2a 2 +2ad (d a) d 2 a 2 + d (d a) a (d a) Φ A λ d + d2 a 2! 2π² 0 a a + d 2 a 2 a ue to the anti-symmetry, the potential on the negatively charged transmission line is Φ Φ +. The voltage between two cylinder is then Φ + Φ 2Φ λ! d + d2 a 2 π² 0 a (4) λ 2π 0 ² 0 ³ d+ (5) d 2 a 2 a 0 / lution

11 Now choose the polar coordinate system as shown (The origin is located at +λ image charge) ρ r +, r ρ 2 2ρ cos φ From e. (2) d 2x 2 d 2 a 2 so r 2 4d2 4a 2 + ρ 2 4 d 2 a 2 ρ cos φ Φ(ρ, φ) λ 2πε 0 0 r r + ³ d+ d 2 a 2 a! ρ 2d 2 4a 2 + ρ 2 4 d 2 a 2 ρ cos φ / lution